Potential

01 拉普拉斯方程和他在笛卡尔坐标系下的解

拉普拉斯方程其人

开局先写一个假设什么东西都不动的麦克斯韦方程组

E=ρϵ0

B=0

×E=0

×B=μ0J

我们知道E=V,把这个带入第一个式子就可以得到著名的泊松equation

2V=ρϵ0

而如果在一个范围内没有任何电荷出现的话 那就可以得到另一个叫做Laplace's equation的东西

2V=0

还挺优雅的 不是嘛

来解方程吧!用一个具体的例子!

这篇文章主要就是要在笛卡尔坐标系下解这个方程,在大二第一学期我们主要使用separation of variable来解,也就是说我们要假设这个V他是等于X(x)Y(y)Z(z)的。我们可以举一个具体的例子来熟悉这种方法

20231013_143402496_iOS

在x = ±b时,V=V0;然后y=0和a的那两个面是接地的。这也就是说Z(z)这一项没有什么用 不妨把他设成1.所以我们现在有

V=X(x)Y(y)

把他带入拉普拉斯方程中

0=2Vx2+2Vy2

就可以得到

Y2Xx2+X2Yy2=0

两边同时除V

1X2Xx2+1Y2Yy2=0

因为第一项只和x有关,第二项只和另一个有关,所以只有这俩都是常数的情况,而且偏微分可以写成全微分

先尝试首项是负数的情况

1Xd2Xdx2=1Yd2Ydy2=k2

这个方程的解为X(x)=A1cos(kx)+A2sin(kx)Y(y)=B1eky+B2eky。我们有边界条件Y(0)=Y(a)=0,也就是说B1+B2=B1eka+B2eka=0,也就是说B1(ekaeka)=0,所以说2B1sinh(ka)=0,也就是说B1=B2=0 这就是说所有地方的V都是0,这显然不是个解,也就是说我们要尝试一下首项是正数的情况

1Xd2Xdx2=1Yd2Ydy2=k2

解是Y(y)=A1cos(ky)+A2sin(ky)X(x)=B1ekx+B2ekx(没错我就换了一下xy,甚至连顺序都懒得换)我们有

Y(0)=0

也就是说A1=0,所以Y(a)=A2sin(ka)=0A2肯定不能等于0了,所以只能让sin(ka)=0,所以说ka=nπ,而A2是随机常数。所以最终我们的解为

V(x,y)=n=0nCncosh(nπxa)sin(nπya)

好家伙 那Cn怎么搞?答案是用X(x)的边界条件

V(b,y)=i=0nCncosh(nπxa)sin(nπya)=V0

我们把左右两边各乘一个sin(mπya),再对y从0到a积分,可以得到

V(b,y)=i=0nCncosh(nπxa)0asin(nπya)sin(mπya)dy=V00asin(mπya)dy

其中

sinnπyasinmπya=12[cos(nm)πyacos(n+m)πya]

所以我们知道只有在n=m时才不是0,其他的时候这个积分等于a2。所以说我们可以得到再m是偶数时,Cm=0,其余时间Cm=4V0mπcosh(mπba)

最终 我们的解就是一个特别恐怖的式子

V(x,y)=4V0πi=0ncosh(2i+1)πxacosh(2i+1)πbaasin(iπya)

 

结束!

02 拉普拉斯方程在球坐标下的解

我们上次介绍了非常优雅的拉普拉斯方程2V=0并且算了算他在笛卡尔坐标系下的一个例子,这次来试一试球坐标

球坐标下的拉普拉斯方程

通过一些矢量微积分,我们可以知道

2V=1r2r(r2Vr)+1r2sinθθ(sinθVθ)+1r2sin2θ2Vϕ2=0

嗯。。。以我现在的知识还是看不出来这有什么优雅的

解方程!

一杯茶,一包烟,一道积分算一天!(虽然我不抽烟)

还是分离变量,假设2=R(r)Θ(θ)Φ(ϕ)

两边同时乘r2RΘΦ,可以得到

1Rr(r2Rr)+1Θsinθθ(sinθΘθ)+1Φsin2θ2Φϕ2=0

左右两边再同时乘一个sin2θ的话,上面这个式子他的左边两项就会只和R和θ有关,而最后一项就会只和ϕ有关。所以我们可以说最后一个项乘个sin2θ是一个常数:

1Φd2Φd2ϕ=K

如果说K>0的话,这个方程的解就是e±Kϕ的线性叠加了,我们在这里其实还是想要一个周期性的解,毕竟ϕ代表的是角度嘛,不能说ϕ=ϕ0+2nπ里面对于任何一个nϕ的解都不一样。所以我们不如去把K设成一个负数,把它叫做m2。这样子的话我们就可以得到

Φ(ϕ)=Ceimϕ

其中的ϕ可以是正数也可以是负数 但必须要是整数。求个导的话就是

d2Φdϕ2=m2Φ

整个的式子就变成了这样子

1Rr(r2Rr)+1Θsinθθ(sinθΘθ)m2sin2θ=0

再找找有没有什么只和一个变量相关的东西?

(没错!)

(是第一项!)

(嘿嘿!)

那我们就可以把第一项设成常数了!

1Rr(r2Rr)=l(l+1)

1Θsinθθ(sinθΘθ)m2sin2θ=l(l+1)

至于为什么把常数设计得这么怪,你之后就知道啦

我们可以把cosθ设为x,那么下面的式子就变成了

ddx[(1x2)dΘ(x)dx]+[l(l+1)m21x2]Θ(θ)=0

苍天啊 大地啊 我都干什么啊!这玩意怎么越写越长啊

别怕,这个方程在之后的数学物理方法中就可以学到 其实叫做Legendre equation。他的解叫做associated Legendre polynomials,可以写成这种形式

Θ(x)=Plm(x)

其中的l是正整数,不然的话这个方程就又不是周期性的了,而且m也只能是从-l到+l

现在知道为啥把常数设成l(l+1)了吧

ΘΦ的积还有一个名字,叫做球谐(spherical harmonic),用Ylm表示

Ylm=CPlm(cosθ)eimϕ

然后就是这个式子了

1Rr(r2Rr)=l(l+1)

求一下导

l(l+1)=2rRdRdr+r2Rd2Rdr2

试一下R=Crx,也就是说

l(l+1)=2Crx1Cxrx1+1Crx2Cx(x1)rx2

l(l+1)=2x+x(x1)

l(l+1)=x(x+1)

所以说x等于l或者l1

所以说R就出来了

R(r)=αrl+βrl+1

把三个东西乘起来 就可以得到

V(r,θ,ϕ)=l=0m=ll(αrl+βrl+1)Ylm(θ,ϕ)

Legendre多项式的一个特点

P有一个很有用的特点,就是他们是正交的,也就是说我们有

11Pn(x)Pm(x)dx=22n+1δm,n

这个东西可以让我们更简单地计算αβ

一个对称的情况

例子 我喜欢例子 现在就让我算例子!!

通常情况下,我们做的题都和ϕ没有关系,所以说我们可以把球谐给写成单独一个关于cosθ的Legendre多项式,也就是说整个的V可以写成

V(r,θ)=l=0(αrl+βrl+1)Pl(cosθ)

我们可以举一个例子:在一个三维空间里面,存在着一个空心导体球在原点,他的半径是a。同时还存在一个沿着z轴向上的匀强电场E,求电势

首先啊,这是一个球,所以我们肯定会用球坐标来解这个方程

其次 我们看一下示意图

Image

这个电势貌似和ϕ没有什么关系!那我们就可以用那个简化版的拉普拉斯方程的解

V(r,θ)=l=0(αrl+βrl+1)Pl(cosθ)

我们首先可以看一看他的end behavior,也就是当r趋近于无穷大时,V应当等于Ecosθ。我们看一下Legendre多项式的前几个方程:

P0(x)=1P1(x)=xP2(x)=12(3x21)P3(x)=12(5x33x)

从中我们可以大概看出来,满足上面的end behavior的情况应该是只有l=1时αβ才不等于0

所以说就可以得到

V(r,θ)=(Er+β1r2)cosθ

还是一样的end behavior

就快要算出来了!

对于β1,我们还有另外一个边界条件没有用,也就是r=a的情况:

V(a,θ)=(Ea+β1a2)cosθ=0

所以说β1=a3E

综上所述,我们能算出来

V(r,θ)=Ercosθ+a3Ecosθr2

可以看出来 其中第一项对应的是电场对应的电势,而后面哪一项就是磁场对应的电势了

 

完事收工!

Formal solution to Poisson's equation

Poisson's equation:

2V=ρϵ0 Has solution V(r)=14πϵ0ρ(r)|rr|dτ

where τ is the volume

This also happens in Magnetism, B=×A, and A is called magnetic vector potential

This is valid because the divergence of B is 0

×(×A)=μ0J

(A)2A

it has gauge freedom that we can choose taht it is divergence free choose A=0

 

2A=μ0J

it looks like three independent Poisson equations

A(r)=μ04πJ(r)|rr|dτ

Example: If we have a one dimentional current i.e. wire, this current will produce

A(r)=μ0I4πdl|rr| B(r)=×A=μ04πdl×(rr)|rr|3 Which is Biot-Savart law!

Multipole expansion

Laplace with azimuthal symmetry:

V(r,θ)=l=0[Alrl+Blrl+1]Pl(cosθ)

where Pl is Legendre polynomial

Apply to a point charge at r

V(r)=14πϵ0q|rr|

(unit charge)

Image

This has azimuthal symmetry, so we can use the above equation

If γ is zero, then cosθ=1, then Pl(cosθ)=1, then

V(r,θ)=14πϵ0q|rr|=l=0Alrl+Blrl+1

Because Pl(1)=1

The expansion would be

1|rr|={1rl=0(rr)lr>r1rl=0(rr)lr<r

Choose the case r<r

The general solutiuon would be

V=14πϵ0qrl=0(rr)lPl(cosθ)

Thus the general solution for r outside a charge distribution is

Image

V(r)=14πϵ0l=0(r)lPl(cos(γ))×ρ(r)dτ

The leading term is the monopole term (l=0)

V(r)=14πϵ0ρ(r)dτ=14πϵ0Qr

Which is proportional to 1r

Next is the dipole term (l=1)

V1(r)=14πϵ0(r)cos(γ)ρ(r)dτ =pr4πϵ0r3

Which is proportional to 1r2

Where p is the dipole moment =rρ(r)dτ

and rrrr=cos(γ)

Then the quadrupole term (l=2)

V2(r)=14πϵ0(r)23cos2(γ)12ρ(r)dτ

Now we can have a nice way of looking at field outside the charges using multipole expansion.

2. Electric polarization

Matter is not free space, it contains atoms

2.1 Bound charges

Electric field can induce a dipole moment in a neutral atom

Image

And the electric moment is defined as p=αE where α is the polarizability of the atom. A list of polarizability can be seen from the handout.

If, for example, the outside E ~ 106 V/m, then the displacement would be about 1015 m, which is much less the size of an atom.

Also, there are polar molecules which already have a dipole moment, but they are randomized because of the present of inner energy KBT

Image

However, an E field can line them up

For a bok of aligned dipoles (per unit volume), define the polarization P as the dipole moment per unit volume as P=np

If they are not aligned, then the polarization is zero.

Question: what is the electric potential produced by a box of dipoles?

Image

V(R)=14πϵ0P(r)R|R|3d3τ=14πϵ0P(r)1Rdτ1R=RR31|rr|=rr|rr|31|rr|=rr|rr|3

Note that (fF)=fF+Ff

(1)V(r)=14πϵ0P(r)|R|dτ14πϵ01RP(r)dτ
(2)=14πϵ0sσbdSR+14πϵ0τρbdτR

where σb=Pn is the surface bounded charge density and ρb=P is the volume bound charge density (the minus sign is because P is defined as the dipole moment per unit volume).

The derivation can be seen from Griffiths' book page 176.

Remember to drop primes!

In bulk, Gauss's Law is E=ρϵ0, but ρ=ρfree+ρbound remember that ρbound=P and hence E=ρfreeϵ0Pϵ0

We define electric displacement

(3)D=ϵ0E+P
(4)D=ρfree

That is much easier, since it only depends on free charges.

The integral form is

(5)Dda=Qfree

Linear dielectrics

For a linear dielectric, P=ϵ0χeE where χe is the electric susceptibility, which is dimensionless.

(6)D=ϵ0E+P=ϵ0(1+χe)E

(1+χe) is the relative permitivity ϵr and it is dimensionless.

Example:

Field from a point charge in a linear dielectric:

(7)sDds=Qfree
(8)D=Qf4πr2r^
(9)E=Qf4πϵ0r2r^

Recap of lecture 5: Polarization: P=np Bound charges: ρb=P, σb=Pn^ Electric displacement: D=ϵ0E+P=ϵ0ϵrE Maxwell: D=ρf Linear dielectric: P=ϵ0χeE Where χe is the electric susceptibility = ϵr1

Question Has D solved everything? -- No. The div equation is easier, but...

Reminder: E=ρtotϵ0 D=ρf And the equation for D is ϵ0E+P=ϵ0ϵrE

(10)×E=0
(11)×D=ϵ×E+×P0

Boundrary conditions

E// is continuous

(12)×E=0
(13)Edl=0

But D// is not continuous (if ×P0 at interface)

D is continuous

(14)D=ρf
(15)Dda=Qf

but E may not be continuous if we have bound charges at the interface

Example

Image

(16)E0//x
(17)E1=E3=E0

D is continuous

(18)D1=D3=ϵ0E0=D2

D2=ϵ0ϵrE2

(19)E2=E0ϵr

So there is a smaller E field in the dielectric

(20)P2=ϵ0(ϵr1)E2=ϵ0ϵr(ϵr1)E0

(P1=P3=0)

Since P2 is constant, ×P=0ρb=0

z=o,σb=Pn=ϵ0ϵr(ϵr1)E0 z=d,σb=Pn=ϵ0ϵr(ϵr1)E0

Example:

There is dielectric sphere, with radius a and relative permittiaty \epsilon_r. The sphere is in a uniform electric field E0. Find V.

We got to use Laplace's equation

General solution to Laplace's equation in spherical coordinates (with azimuthal symmetry) is

V(r,θ)=l=0(Alrl+Blrl+1)Pl(cosθ)

where Pl is the Legendre polynomial.

Solve separatively for r > a and r < a

As r, VE0rcosθ As r0, V should be finite (not blow up) Bl=0 for r<a

case r<a: V=Vinl=0AlrlPl(cosθ)

And for r>a: V=Vout=E0rcosθ+l=0Blrl+1Pl(cosθ)

As for r = a, E//,D,V are continuous

l=0AlrlPl(cosθ)=E0acosθ+l=0Blal+1Pl(cosθ)

Al={Blal+1l1E0+B1a2l=1

E0=1rVθ is continuous

generates same expressions

 

That is not terribly useful, so we will try another one

D=ϵ0ϵrVr is continuous

ϵ0ϵrVr|r=a=ϵ0Voutr|r=a

ϵrl=0lAlal1Pl(cosθ)=E0cosθ+l=0(l+1)Blal+2Pl(cosθ)

Al={1ϵrl+1lBla2l+1l11ϵr(E02Bla3)l=1

Hence, we got two expressions for A in terms of l. They cant both be right unless Al=Bl=0 for l1

On the other hand, we have

A1=E0+B1a3=1ϵr(E02B0a3)

ϵr()=0

Hence

(1ϵr)E0+(ϵr+2)B1a3=0

B1=a3E0(ϵr1ϵr+2)

A1=3E0ϵr+2

Now, put everything together

V={3ϵr+2E0rcosθuniform fieldr<aE0rcosθuniform field+(Er1Er+2E0a3cosθr2)dipole fieldr>a

Interestingly, we all have uniform fields applied inside and outside. And if in vacuum, we will have uniform field outside (dipole field vanishes)

 

P inside sphere:

P=ϵ0(ϵr1)Ein

where Ein is 3E0ϵr+2

Dipole moment of sphere

p=43πa3P=4πa3ϵ0(ϵr1)E0ϵr+2

Polarizability

α=PE0=4πa3ϵ0ϵr1ϵr+2

Bound charge on the surface

σb=Pn^=Pcosθ=3ϵ0(ϵr1)ϵr+2E0cosθ

3 Magnetization

3.1 Current Loop

From the handout, we have the following vector identity:

r^rdl=(dS)×r^

Start from Poisson's equation:

2A=μ04π(J(r)|rr|)dτ

Current loop:

A(r)=μ0I4π1|rr|dl

Image

recall the expansion

1|rr|=1rl=0(rr)lPl(cosθ)

A(r)=μ0I4πrl=0(rr)lPl(cosθ)dl

l = 0 term is zero

A0(r)=μ0I4πrdl=0

n = 1 term is the next most important term

A1(r)=μ0I4πr(1r)rcosθdl

and rcosθ=rr^ hence

A1(r)=μ0I4πr2rr^dl=μ0I4πr2m×r^
where m=IsdS=IS

Is the magnetic dipole moment of the current loop.

Lets stop the expansion before we get too complex stuff.

And that looks like a dipole right?

Put mz in spherical coordinates:

A1(r)=μ04πr2msin(θ)ϕ^

B(r)=×A=μ04πr3msin(θ)r^μ04πr2mcos(θ)θ^

which is same as electric dipole with P4πϵ0μ0m4π

Image

Field patterns look the same at large r, but at small r, the magnetic field is not singular.

3.2 Magnetic Properties

In a field B,a magnetic material will acquire a magnetization M=nm where n is the number of magnetic dipole and m is the magnewtic moment of one atom

There are three main effects:

  1. diamagnetism MB and that effect is very weak

All materials exihibits it!

  1. Para-magnetism M+B and it's stronger

That is often shown in materials with unpaired "spins" Image (MB but only at small B) Example: CuSO45H2O

  1. Ferromagnetism

Very strong QM effect E.g. Fe, Co, Ni It is a non-linear effect of M(B)

Recap to lecture 7:

Current loop A(r)=μ04πr2m×ϕ^ Magnet moment m=IS where S is the vector area of the loop Magnetization M=nm where n is the number of magnetic dipole and m is the magnewtic moment of one atom

3.3 Field due to a box of magnetic dipoles

When doing this, we start with some vector identities:

start with Divergence theorem:

Vudτ=Sudτ set u=v×c where c is a constant vector

LHS: sv×cdS=csv×dS=0 because v×dS is a vector perpendicular to c

RHS: τ(v×c)dτ=τc(×v)dττv(×c0)dτ

Therefore, for any c, Sv×dS=τ×vdτ

Image

Assume there is a box of magnetic dipoles, and there is one dipole at position r

A(r)=μ04πM(r)×(rr)|rr|3dτ

which equals

mu04πM(r)×1|rr|dτ

We have

×(Mrr)=(1|rr|×MM×1|rr|)

Hence

A(r)=μ04π×M(r)|rr|×M(r)|rr|dτ

The last term can be transformed into a surface integral:

A(r)=μ04π×M(r)|rr|dτ+μ04πM(r)×dS|rr|

This looks like μ04πτJ(r)|rr|dτ+μ04πsKb(r)|rr|dS

Where J=×M is the bulk bound current density and Kb=M×n^ is the surface bound current density.

Remember to drop the prime!

Having done this, we can fix up Amperes law:

Amperes law: ×B=μ0(Jf+Jb)

Where Jf is the free current density where you can connect or do sth like that; and Jb is the bound current density, it can be written as the curl of M

Hence the expression for free current would be Jf=×(Bμ0M)

Stuff in the bracket is defined as the magnetic field H and hence the free current density has a simpler form

Jf=×H

Its integral form is

Hdl=If

However, B=0 becomes H=M

The equtions to remember would be

D=ϵ0E+P B=μ0(H+M)

Linear materials

M is proportional to B M=χmH where χm is the magnetic susceptibility

B=μ0(H+M)=μ0(1+χm)μrH=μ0μrH

where μr=1+χm is the relative permeability

If willing to be strict, H would be called magnetic field strength and B would be called magnetic flux density

3.4 Boundary conditions

one is not changed:

Image

B=0B1=B2

Making this cylinder flater and flater, we can see that the B field is continuous

However, H1=H2 is not true because the H=M

Image

×H=JfH1H2 (Assuming there is no surface free currents)

But B is not continuous because you can have bound surface currents at interface and ×B=μ0(Jf+Jb)

E,D,H,Bare continuous

Assuming there is no surface free charges/currents

3.5 Magnetic scalar potential

If Jf=0 everywhere, then

×H=0H=ϕm

where ϕm is the magnetic scalar potential

If, we are dealing wit linear material,

B=μ0μrϕm

If, in addition, H=0

Then we can use the Laplace equation to solve for ϕm

3.6 Ferromagnetism

One way to define magnitization is = limδv0iδvmiδv Microscopely, ferromagnet has M 0, even in B = 0 The reason for ferromagnetism is the (what will be learnt next term), the exchange interaction Only in certain materials: Fe, Co, Ni, Gd

There is an energy cost due to the stray field of the magnetic dipoles.

Image

From the picture having stray field, it can be seen that on the tip and bottom of the field, there is a divergence of magnitization, leading to divergence in H. Meaning that field spreads out whenever you have magnetic moment going into the surface and not flowing out.

The energy cost would be

B22μ0dτ It is energetic flavorable to form domains

Image

This will cost low energy since there is no stray field and hence no divergence in H. (Average over the surface, M is zero)

Image

When applying a B field, the correlated domain is expanded and finally they are all aligned.

So, the magnetization process involves moving domain walls.

The process is highly non-linear hysterisis loop Image States that is interesting

  1. Saturation (right up)
  2. Remanence (middle up)
  3. Coercivity (left middle) Better picture is in handout

The field which is positive but M is zero is called coercive field.

We can modle this as

M=(μr1)H only if μr=μr(H)

and μr(H) is a multivalued function (because it depends on histroy)

Hard materials

Hc, Mr are large

Soft materials

Hc, Mr are small

Example: Magnetization of a ring Iron ring, radius r, and current I with N terms used to magnetize the ring Image

Have a Ampere loop around the ring, then use Hdl=NI Since H only depends on the free current, it is favorable for us to use it.

Continue the calculation

H2πr=NIH=NI2πr

Hence B = μ0μrH=μ0μrNI2πr

Now lets put a gap in the ring

Image

The cutout, x, would much less than r

Lets use the same trick, ampere's law

Hdl=NI

Hcore(2πrx)+Hgapx=NI

B Is continuous, Bcore=Bgap Substitute in the function Bcoreμ0μr(2πrx)+Bgapμ0(x)=NI

Rearrange that Bgap=μ0μrNI(2πrx)+μrx

If μr>>2πrx, then Bgap becomes about μ0NIx

Is could be very large if x is small. But for that condition to work, you need μrNI to be very large.

4 Electromagnetic waves in material

Let's let everything move!

4.1 Displacement current

Conservation of charge

(21)Js=ρft

However, this is incompatible with

(22)×H=Jf

Lets take the divergence for both sides, we get

(23)(×H)=0=J+tDρf

add an additional term to the current density Jf=×H

(24)×H=Jf+Dtdisplacement current

LHS: ×(Bμ0M)=1μ0×B×MJb

Where Jb is the bound current density.

RHS =Js+Dt=Js+Pt+ϵ0Et

we could write ×B=μ0(Jf+Jb+Jp)+μ0ϵ0Et

where Jp is the polarization current density. which equals Pt

Note that Jp=ρpt from conservation of charge.

Thus the polarization current responds to changes to bound charge, and hence in P

4.2 Maxwell's equations in insulating linear dielectrics

Since it is insulating linear dielectrics, we have Jf=0 and Jb=0

Hence, we could get Maxwell's equation

(25)D=0×E=BtB=0×H=Dt

remember, D=ϵ0ϵrE+P and B=μ0μrH

which gives

(26)E=0×E=BtB=0×B=μ0μrϵ0ϵrEt

Consider

(27)×(×E)=(E)2E=t×B
(28)2E=μ0μrϵ0ϵr1v22Et2

which is wave equation.

v=cn where c = 1ϵ0μ0 and n = ϵrμr where n is also called refractive index

Plane waves solutions

Lets choose propagation parallel to z, and hence Ψx=Ψy=0

remember that E=0Ezz=0 similarly, B=0Bzz=0

we also have ×E=BtBzt=0

And ×B=μ0μrϵ0ϵrEtEzt=0

Hence, Ez and Bz are constant in z and t, they are not part of wave motion

now analyze the x,y components of curl:

(29)Eyz=Bxt,Byz=1v2Ext

Ex,By are solutions

Lets then take

(30)2Ext2=1v22Ext2
(31)Ex(z,t)=Ex0ei(±kzωt)x^

Then, we could have

(32)B(z,t)=B0ei(±kzωt)y^

And then we could get the wave travelling in ±z direction

kE0=ωB0and ±kB0=ωv2E0E0B0=±ωk=±v

Define Impedance Z as

(33)Z=|E0H0|=μ0μrϵ0ϵr

remember that H0=B0μ0μr

The motivation of doing so is that v=Edl and I=Hdl

So dimension would work

For free space, then, ϵr=μr=1 and Z=μ0ϵ0=377Ω

Remember that ×E=B˙, and use E, B ei(krωt), we could get

ik×E=(iω)B=iωμ0μrHZ=|EH|=μ0μrωk

Which gives the same answer because v=cn=ωk=1ϵ0ϵrμ0μr

which is this wave

Image

4.3 conductors

Remember that Jf=ρftJp=ρptJb=0 The last one is 0 because (×M)=0

For conductors, we have

ρf=0 since there are no free charges in equilibrium Jf=σE from Ohm's law where σ is the conductivity D=ϵ0ϵrE and B=μ0μrH from linearity

Then we could get Maxwell's equation in conductors

E=0×E=BtB=0×B=μ0μrσEconduction J+μ0μrσ0σrEtdispalcement J

Free charge will decay to zero in a short time τ, and it is easy to prove (said Blundell)

J=ρt

Where J is equal to σE from Ohm's law

and E is equal to ρϵ0ϵr from Gauss's law

ρ(f)=ρ(0)etτ

Where τ=ϵ0ϵrσ

If the metal has great conductivity, then τ is very small, and hence ρ(f) is very small.

Let's consider the electromagnetic wave having frequency ω, so we would like to compare 1ω with τ:

ConditionConductor TypeCharge ResponseConductivity
ωτ1Good conductorCharges respond very quicklyσϵ0ϵrω Conduction current dominates
ωτ1Bad conductorCharges respond very slowlyσϵ0ϵrω Displacement current dominates

Take real life examples

 σ(Ωm)ϵrσϵ0ϵr(S1)
metal10711019
Silicon410411.7107
Glass1010510

Note that visible light would have frequency ~ 51014 Hz

Let's now do some electromagnetism

×(×E)=(E)02E=t×B2E=μ0μrσEt+μ0μrσ0σr2Et2

Again, this would yield transverse plane waves with E,B to each others

E=E0x^ei(k~zωt)

with k~=k+iκ

k~2=iμ0μrσωμ+μ0μrμϵ0ϵrϵω2

(k+iκ)2=k2κ2+2ikκk2κ2=μ0ϵ0ω22kκ=μσωk=μσω2κ0=(μσω2)21κ2κ2μϵ0ω20=(κ2)2+μϵω2(κ2)(μσω2)2κ2=μϵω22±(μϵω22)2+(μσω2)2κ2=μϵω22[±1+(σϵω)21]

Taking the positive root κ=μϵ2ω1+(σϵω)21

Sub into original equation

k=μσω2κ=μϵ2ω1+(σϵω)2+1 E=E0x^eκzezδei(krωt)

Where δ=1κ is the skin depth

Reminder: Good conductors have σϵω

k=κ=μϵ2ωσϵω=μωσ2

We could therefore have

2E=μσEt+μϵ2Et2

We could also neglect the last term, since 2Et2Et

k~2=iμσωk~=1+i2μσωk~=κ=μσω2

Hence, we could have

δ=1k=2μσω

For a typical metal, δ is

{few nm - visible lightfewμm - microwavefew mm - radio waves

Lets go to poor conductors

Poor conductors has σϵω, hence

kμϵ2ω2=μϵω κ=μϵ2ω(1+12(σϵω)2+1)12

which equals to σ2μϵ which is independent to ω

In an insulating dielectric, σ=0, hence κ=0, and hence k=ωv as expected.

Lets return to previous equation

Lets consider the curl equation in the conductor

Exz=Byti(k+k)E0=iωB0z=μE0B0=μωk+iκk~=μωk2+κ2eiϕ

If we expand, ϕ would be

ϕ=tan1(1+(Q/ϵω)211+(Q/ϵω)2+1)12

For a good conductor, σϵω, hence ϕtan11=π4

So this means that B lags behind E in a metal

4.4 Poynting vectors

Work done on charge

δq=ρ=δτ

δF=δq(E+v×B)δFdl=δq(E+v×B)vδt=EJfδτδt

where Jf eaquals to ρv

Rate of work on charges F=dwdt=EJfdτ=ddtumechEnergy densitydτ

From Maxwell's equation: Jf=×HDt We have

EJf=E×HEDt

By dotting everything, and then

(E×H)=H×EE×H EJf=H×EBtEDt(E×H)

Where we call H×EBtEDt "tuEM" (remember that uEM=12(BH+ED) which equals to energy stored in EM field per unit volume) and (E×H) as "S=E×H" or Poynting vector.

How is this working?

Assume that we are using a linear media: EDt=12(ED)tHBt=12(BH)t Remember that 12ϵ0E2=12DEB22μ0=12BH}In free space

Bringing everything together, we could get ddt(umech+uEM)+S=0

Where S is the poyting vector, or equiviantly,

ddt(umech+uEM)dτ+Sda=0

We could say that, therefore S is the energy flux density, or the rate of flow of energy per unit area in the direction of S.

Example: a capacitor

Image

The stored energy increase at rate

U˙=QCdQdtU=Q22C

also: Image

Hence, we have

U˙=S2πrd=QcdQdt

where S=EH=QdcdQdt2πr

There is another example

Image

HSπa=IE=VlS=VlI2πaSda=IV=I2R

4.5 Radiation pressure

EM waves are made up of photons, and hecne they have momentum

E=pc

Transport of energy is appoinated by transport of momentum

Prad=<S>c

For a perfect absorber, where Prad is the radiation pressure

Example For a plane EM wave in free space, we have

U=12ϵ0E2+12B2μ0but E=cBU=ϵ0E2E=12E0cos(kzωt)x^=E0cos2(kzωt)E2=12E02u=12ϵ0E02S=12ϵ0E02c=I Where I is the intensity of wave Prad={12ϵ0E02perfect absorberϵ0E02perfect reflector

Sunlight: I~1kWm2 Prad=105Pa FYI, Patm=105Pa

Example Consider a star which is growing by accretion

i.e. matter is falling onto it uniformly in all directions

The star has luminosity L (e.g. Lsun=4×1026w)

Energy flux = L4πR2[Wm2]

Radiation pressure: L4πR2C

Outward:

Force/unit mass = kL4πR2C where k is the opacity, which is area/unit mass, which is a constant

Inward: Force due to gravity/unit mass = GMR2

Since they balance, we have

L=4πCGMk

which is called Eddington Limit (Upper limit of luminosity of stars that accrete (isotopically))

4.6 EM waves - reflection and refraction

Image

Left: Eiei(k1zωt)+Erei(k1zωt)

Right: Etei(k2zωt)

Using electromagnetic boundary conditions, we could get

E is continuous Ei+Er=Et

H is continuous EiZ1ErZ1=EtZ2

Putting two equations together

ErEi=Z2Z1Z2+Z1EtEi=2Z2Z2+Z1

Where Z=μϵ

|Poyting vector| =S=|E×H|=E2Z

We expect Sincident=Sreflected+Stransmitted

Where they equal to

Ei2Z1+Er2Z1=Et2Z2

separately

Lets now have angles Erei(krrωt)

Image

Eiei(kirωt)

Etei(ktrωt)

Choose ki in x-z plane

At z = 0, E is continuous and this holds for all x y and t ω must be the same

kir=krr=ktr for all x, y at z=0

Take r=(0,y,0)

ki,kr and kt all lie in the xz plane (the plane of incidence)

Take r=(x,0,0) so kir=ksinθx

|ki|=|kr|=k1|kt|=k2k1sinθi=k1sinθrθi=θr, law of reflection=k2sinθt

Remember that ωk=cn, And the last two would lead to

sinθtsinθi=k1k2=n2n1law of refraction, or snell’s law

Where n=ϵrμr

Fresnel equations

Worring about polarization directions

We work in those steps

  1. E in the plane of incidence

"parallel-like" = parallel

Remind that E,H and k form a right-handed system

Image

 incidentreflectedtransmitted
ExEicosθiErcosθrEtcosθt
EzEisinθiErcosθrEtsinθt
HyEiZ1ErZ1EtZ2

E continuous Ex continuous Eicosθi+Ercosθi=Etcosθt

ErEi=Z2cosθiZ1cosθiZ2cosθ1+Z1cosθiEtEi=2Z2cosθiZ2cosθi+Z1cosθi

Now look for Fresnel equations for p-polarizations

  1. E perpendicular to the plane of incidence

"s-like" s = senkrecht = perpendicular

Image

 incidentreflectedtransmitted
EyEiErEt
HxEiZ1cosθiErZ1cosθrEtZ2cosθt
HzEiZ1sinθiErZ1sinθrEtZ2sinθt

E continuous EycontinuousEi+Er=Et H continuous HxcontinuousEiZ1cosθi+ErZ1cosθr=EtZ2cosθt

Remember that

Z=μrμ0ϵrϵ0=Z0nn=ϵrwhile μr=1

Let's set μr=1 Then, Z=μrμ0ϵrϵ0=Z0nn=ϵr

So we can replace Zi with 1ni in expressions involving ratios of Z's.

e.g. Fresnel equations for p-polarization

r=ErEi=1n2cosθt1n1cosθi1n2cosθt+1n1cosθi =n1cosθtn2costhetain1cosθt+n2cosθi

Use Snell's law n1sinθi=n2sinθt r=sin2θtsin2θisin2θt+sin2θit=4sinthetatcosθisin2θt+sin2θi

For s-polarization, we have

r=sin(θtθi)sin(θt+θi)t=2sinθtcosθisin(θt+θi)

We also have

n1sinθi=n2sinθtsinθt=n1n2sinθicosθt=1sin2θt=1n12n22sin2θi

Fresnel equations:

n1cosθi on top and bottom

α=cosθtcosθi=1cosθi1(n1n2sinθi)2β=n2n1

 E(p)E(s)
rαβα+β1αβ1+αβ
t2αα+β21+αβ

Remember, EM waves have an energy flux given by

S=|E×H|=E2Z

Intensity coefficients

R=IrIi=|r|2={(αβα+β)2(p)(1αβ1+αβ)2(s)

T=ItIi=|t|2n2n1cosθtcosθi

where n2n1 is due to waves at ifferent speeds. And cosθtcosθi is due to the wavesfronts at different angles.

T=|t|2αβ={αβ(2α+β)2(p)αβ(21+αβ)2(s)

Lets check in certain cases

rs=rp=1β1+β=n1n2n1+n2ts=tp=21+β=2n1n1+n2Rs=Rp=(n1n2n1+n2)2Ts=Tpn2n14n12(n1+n2)2=4n1n2(n1+n2)2

Example: air/glass interface

n1=1n2=1.5rs=rp=0.2Ts=Tp=0.8Rs=Rp=0.04Ts=Tp=0.96 4% of light is reflected, 96% is transmitted

[If n2, for example, = 1.75, Rs=Rp=0.074, which is sa problem]

Lets take another go at differnet angle

sinθi=1cosθi=0 α rp=1rs=1tp=0ts=0

At θc=sin1β, α=1cosθc1(sinθcβ)2=0

Hence, we have rp=1rs=1 tp=2βts=2

sinθc=β

rp vanishes at the certain angle called "Brewster's angle" θB rp=αβα+β=0 when α=β=n2n1 1cosθB1(sinθBβ)2=βsquare both sides:sec2θB1β2tan2θB=β2(11β2)tan2θB=β21(β21)β2tan2θB=(β21)tanθB=β

Method of producing polarized light

Image

Reflectted light polarized with E perpendicular to the plane of incidence

Polarizing sunglasses with transmission axis vertical reduce glare because reflected light is mainly horizontally polarized.

Total internal reflection

n1sinθi=n2sinθtsinθt=sinθin2/n1

Since n2<n1, there will be total internal reflection when sinθt>1

For transmitted wave, its like

Lets now consider the Plane travelling wave Ex=E0ei(kzωt)×E=B˙=iωBB=1iω|i^j^k^xyzEx00|=1iω(0zExyEx)=1iω(0ikEx0)=Exω/ky^=Excy^

For s-polarization, θi>θc, we have

sinθt=sinθin2/n1>1cosθt=isin2θt1rs=n1cosθin2cosθtn1cosθi+n2cosθt=AiBA+iBRs=|rs|2=A2+B2A2+B2=1ts=2n1cosθin1cosθi+n2cosθt=2AA+iB

So is complex Et=y^tsEiei(k2[xsinθtωt])ez/δ

where δ=λ22π1(sin2θt1)12 ×E=B˙=iωBB=1iω|i^j^k^xyz0Et0|=1iω(zEt0xEt)=(iEtωδ0k2sinθtEt/ω)

Bz is in phase with Ey transport of energy along x Bx is out of phase with Ey no transport of energy along z <S>=<E×H> H is the complex conjugate of H

Reminder on conductors ×(×E)=(E)2Et×Bμ0μrσE+μ0μrσ0σrEt=0if good conductor2E=μ0σEtk~2=iωμ0σk~=k+iκk=κ=μ0σω2×E=B˙Byt=(×E)y=ExziωB0==i(k+iκ)E0Z=E0H0=E0B0/μ0=μ0ωk+iκ=2μ0ωσ11+iZ=μ0ω2σ(1i)

Reflection from a meatal surface

airmetal
Z1=μ0ϵ0Z2=μ0ω2σ(1i)

|z2||Z1| for a good conductor because σϵ0ϵrω

write α=μ0ω2σμ0ϵ0=ωϵ02σ1

r=ErEinormal incidence=Z2Z1Z2+Z1=Z2/Z11Z2/Z1+1=α(1i)1α(1i)+112α+1+2α+=14α+O(α2)=12ωδc+O(α2)=14πδλ+O(α2)

where δ=2ωμ0σ is the skin depth most of the EM wave intensity is reflected metals are shiny!

4.7 Plasmas

Plasmas are neutral gas of charged particles, such as ions and free electrons (like metals)

Examples: where you can find plasma

Its density varies from 1021035kg/m3, temperature is 1001013k

we would only focus on cold plasma

Lets consider a slab of plasma

Image

having number density n

the positive ions are fixed in place, and now lets move electrons by distance ξ

E field: E=σϵ0 and σ=neξ

mξ¨=eEwhereE=neξϵ0ξ¨+ωp2ξ=0which is SHMwhere ωp2=ne2ϵ0m

This is the SHM at the plasma frequency ωp

Now, lets drive charges with EM wave Eeiωt (we could ignore B if vc)

Hence, we could get ξ=ξ0eiωt

mω2ξ0=eE P=neξ0=ne2mω2E=(ϵr1)ϵ0Eϵrn~2=1ωp2ω2

where n~2 is the refractive index, and that can be imaginary

we could plot the relation between ω and ϵr

Image

at ω>ωp, n~ is real, hence EM waves can propagate

at ω<ωp, n~ is imaginary, hence EM waves cant propagate

For example, metals are shiny, but only at optical frequencies. They will transport if it is going to much shorter wavelengths.

e.g. ionosphere

AM radios would be refracted + reflected (~ 1 MHz)

FM radio and TV radios would escape (~100MHz)

Lets look at the dispersion relation again

ωk=cn

we could get

n~=1ωp2ω2=c2k2ω2ω=c2k2+ωp2

The dispersion relationship would be like

Image

Hence, we could conclude that

We have 2ωdω=2c2kdk

dωdk=c2kω=c1ωp2ω2

We could have end behaviors:

vg0asωωpvgcasω

Let's then take

E=E0ei(krωt)

Choose k to be (0,0,k), we have

P=ne2mϵ0ω2(ϵ0E)=(ωpω)2ϵ0E

We could hence get Maxwell's equation

×H=D˙t×E=Bt=μ0Ht×(×E)=μ0t×H=0μ02t2DLHS=(E)2E=μ02t2(ϵ0E+P)

For the wave, we have

ik tω2E k2Ek(kE)k2Ezz^=ω2ωp2c2E

Transverse solutions:(100) (010) we have ω2=ωp2+c2k2

As for longitudinal solutions, we have (001), we have, hence, ω2=ωp2

We can therefore classify waves in the ω k graph

Image

4.8 Dispersion

Refractive index changes with frequency

model electrons as a classical damped oscillator mx¨+mγx˙+mω02x=qE0eiωt

Assume that x=x0eiωt (ω2iωγ+ω02)x0=qE0mP=nqx0=nq2mE01ω02ω2iωγ=ϵ0(ϵ~r1)E0ϵr~=1+nq2E0ϵ0m1ω02ωiωγn~2=ϵ~rFor a gas, ϵ~=1+small quantityϵ~12=1+12small quantities+Re(n~)=1+nq22ϵ0mω02ω2(ω02ω2)2+ω2γ2Im(n~)=nq22ϵ0mγω(ω02ω2)2+ω2γ2

Image

As ω0, Re(n~)1+nq22q0ω02, Im(n~)0 As ω, Re(n~)1, we could get a better illustration, therefore, for the real part and imaginary part of n~

Image

We can also conclude that Im(n~) corresponds to the absorption of light, and Re(n~) corresponds to the refraction of light and generally increases with frequency

The sharp drop in the real part of n~ in the real part of n~ near ω0 is called anomalous dispersion

5. Confined EM waves

5.1 Transmission lines

An example of guided wave

Lets think in a super long circuit

Image

If the time for signal to transverse the circuit is not 1ω, we need to consider the wave behavior of the signal.

Image

We could have

Remember: we are defining capacitance as C per unit length, and inductance L per unit length

Q=(Cδz)vdQdt=Cδzvt=I(z,t)I(z+dz,t)=δz(Iz)Iz=Cvt equation 1Φ=(Lδz)IdΦdt=LδzIt=V(z,t)V(z+δz,t)=δz(Vz)Vz=LItequation 2

If we plug equation 1 into equation 2, we get

2Vt2=1LC2Vz2

For wavelike situations, we have

V(z,t)=f(zvt)+g(z+vt)

where f and g are arbitrary functions, and v is the wave velocity which is 1LC.

Note: the wave velocity is not the speed of light, but the speed of the wave in the circuit.

From equation 1, wee could have dVdI=1Cdzdt=1C1LC=1LC The impedance would be, then Z=±LC where thr ± is the direction of the wave.

Instantaneous power

=V(12LI2+12CV2)energy stored/length=12I2z+12V2z=I2z=LCz=IV

Example

We have coaxial transmission line, which has

C=QV

Remember: C is per unit length

Image

We would have

E2πrl=Qϵrϵ0

Image

Hence, we have

V=baQ2πϵ0ϵrdr=Q2πϵ0ϵrlnbaC=2πϵrϵ0lnbaB2πr=μrμ0IΦ=fracΦI=μrμ02πlnba

Where, remember, Φ is the flux per unit length

σ=1LC=1μrμ0ϵrϵ0Speed of light in dielectricZ=LC=μrμ0ϵ0ϵrlnba2π

Then, a common type of transmission line:

Example: Strip transmission line

Image

ad so we could ignore edge effect

We have

C=ϵ0ϵradL=μ0μrdaσ=1LC=1μ0μrϵ0ϵras beforez=LC=μ0μrϵ0ϵrda

Boundrary between two transmission lines

Image

For x<0, we have

V1=Viei(ωtkx)+Vrei(ωt+kx)

For x>0, we have

V2=Vtei(ωtk2x)

At x = 0, we need to match voltages

Vi+Vr=Vt(1)

Match currents, we have

ViVrZ1=VtZ2(2)

From (1) and (2), we have r=Z2Z1Z2+Z1, t=2Z2Z2+Z1

As for power, we would have Vi2Z1=Vt2Z2+Vr2Z1 as expected.

Termination of load

Image

VTIT=ZT is the boundary condition

r=ZTZcZT+Zct=2ZTZT+Zc

We would also like to consider special cases

 ZTrtVTIT
Short circuit0-1002Ii
Open circuit122Vi0
MatchedZc01ViIi

When ZT=Zc, we would get maximum power transfer

proof

Incident power =Vi2Zc=12Vi2Zc

Power on the load =VT2ZT=12VT2ZT=124ZTVT2(ZT+Zc)2(3) Hence, Power transmittedincident power=(3)(4)=4ZcZT(ZT+Zc)2

ddZT[ZT(ZTZc)2]=0(ZT+Zc)212(ZT+Zc)ZT(ZT+Zc)2=0(ZT+Zc)(ZT+Zc2ZT)=0ZT=Zc

Input impedance of short sections

Image

Input impedance

Zin=V(l)I(l)=Vieikl+VreiklViZTeiklVrZceikland Vr=rVi wherer=ZTZcZT+ZcZin=(ZTcos(kl)+iZcsin(kl)Zccos(kl)+iZTcos(kl))ZcFor a λ2 line (kl=π)Zin=ZTFor a λ4 line (kl=π2)Zin=ZcZT

If ZT = 0 Zin=iZctan(kl) If ZT = Zin=iZccot(kl)

For a λ4 line, we have Zin=Zc2ZT

Image

Choose Zc2ZT=Zc i.e. Zc=ZcZt The transmission line has been perfectly terninated and there is no reflected wave

Waveguides

Example of interfering EM waves

Consider 2 EM waves with Ey, travelling along k±=k0(±sinθ,0cosθ)

And we add them up

Image

E=E02(ei(k+rωt)ei(krωt))y^=E02ei(k0zcosθωt)(eik0xsinθeik0xsinθ)y^=E0iy^ei(k0zcosθωt)sin(k0xsinθ)

And so it is zero when k0xsinθ=nπ i.e. x=na when aa=πk0sinθ

Nodal plane would be like

Image

We could, therefore, insert conducting sheets at places where E=0

This demonstrates (at least in 1 dimension) that guided waves are possible

When confined

Image

B=1cE02ei(k0zcosθωt)[eik0sinθ(cosθ0sinθ)+eik0sinθ(cosθ0sinθ)]=E02cei(k0zcosθωt)(icosθsin(k0xsinθ)0sinθcos(k0xsinθ)) only E is transverse, B is not

Rectangular waveguide

Image

Rectangular cross section metal walls

E, Beiωt E=B=0 inside conductor walls B and E are continuous B and E are zero at the walls

Consider Maxwell's equations inside waveguide E=0×E=iωBB=0×B=iωc2E

Hence, using our old method, we could have ××E=iω×B=ω2c2E Where the LHS would be (E)02E=2E, so we can have a second order differential equation

2E+ω2c2E=0

Which is called the Helmholtz equation

One set of solution of this equation have E transverse (called TE modes) Ez=0 Since E = 0 at walls, we could get {Ex=0at y=0,bEy=0at x=0,a And since E(r,t)ei(kgzωt), we would like to put it in Helmholtz equation [2x2+2y2+(ω2c2kg2)]E=0

Notice that Ez = 0

Boundary conditions ExsinnπybEysinmπxaEx=f(x)sinnπybei(kgzωt)Ey=g(y)sinmπyaei(kgzωt)

From E=0, we have Exx+Eyy+Ezz0=0, hence we could continue

fxsinnπyb+gysinmπxa=0fx=sinmπxaf(x)=amπcosmπxagy=sinnπybg(y)=bnπcosnπyb

If m = 1, n = 0, we would call it TE10 mode

Ex=Ez=0Ey=Aπasinπxaei(kgzωt)Bx=Akgωsinπxaei(kgzωt)Bz=iAω(πa)2cosπxaei(kgzωt)

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We could have Helmholtz equation for Bz as well ω2c2kg2(mπa)2(nπb)2=0 ω2c2=kg2+(mπa)2+(nπb)2kc2 Where kc is the cutoff wave vector

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There is a minumym frequency allowed allowed by the waveguide

2ωdωc2=2kgdkgVgroup=dωdkg=c2kgω=ckgkg2+kc2vphase=ωkg=ckg2+kc2kg

And vgroupvphase=c2 kg=k0cosθkc=k0sinθωc=k0

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We can, therefore, consider the TE10 mode ad an EM wave bouncing off the walls at angle θ0 with respect to the walls

 

6. Special relativity and electromagnetism

Just a brief introduction, no worries.

We have spacetime four vector (ct,r), who has dot product with itself as

xμxμ=c2t2r2τ2=t2(1x2+y2+z2c2t2)=t2(1v2c2)τ=tγγ=(1v2c2)12t=γτtime dilation

We also have momentum four vector (Ec,p), who has dot product with itself as

pμpμ=E2c2p2=m2c2sinceE2=p2c2+m2c2(m=rest mass)

We also have differential operator μ which is written as

μ=(1ct,)

We would just introduce this here

Then, we could have current density four vector

Jμ=(cρ,J)

We would consider charge at rest, then

Jμ=(cρ,0)

This could be "boosted" to a moving frame with speed v

Jμ=(ρc,J)

Where ρ=γρ0 due to length contraction and J=γρ0v

Then, we could have continuity equation

ρt+J=0μJμ=0

In lab frame S, then, we could have

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remember that the wire itself has no net charge

λtot=+λλ=0

In the test charge frame S', we could have

Hence γtot=γ+λγ(λγv)

In the rest frame of negative charges, they have charge density of λγv

λtot=γuλλγu(1+uvc2)=λγuuvc2E=λtot2πϵ0r=λuvγu2πϵ0c2rF=qE=λuvγuq2πϵ0c2rF=Fγu=(λv)uq2πϵ0c2rnote thatλv=I=qu(μ0I2πr)note thatμ0I2πr=B=quBWhich is lorentz force

Lets remember Maxwell's equations

B=×AE=VAtIfVVχtandAA+χWhere χ is a scalar fieldAμ=(Vc,A)Aμ=(Vc,A)AμAμμχgauge transformation

We would, therefore, need a new object called the field strength tensor

Fμν=μAννAμWhere μ,ν are indicesFμνFμνμνχ+νμχ=FμνFμν=(0ExcEycEzcExc0BzByEycBz0BxEzcByBx0)Fμν=(0ExcEycEzcExc0BzByEycBz0BxEzcByBz0)

And the function μFμν=μ0Jν gives us the four maxwell's equations. The first half would be simple to deduce, but the second half would be a bit more complicated so we are not going to do it here.

First two:

iFi0=μ0J0E=ρϵ0μFμi=μ0Ji1c2E+×B=μ0J